OFFSET
1,2
COMMENTS
a(n) is divisible by 264 = (2^3)*3*11 = A006863(5), except for n = 1, 2, 3 or 11. See Lugo. - Peter Bala, Jan 13 2024
REFERENCES
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 199, #3.
LINKS
Robert Israel, Table of n, a(n) for n = 1..10000
Michael Lugo, A little number theory problem (2008)
Wikipedia, Jordan's totient function.
FORMULA
a(n) = Sum_{d|n} d^10*mu(n/d).
Multiplicative with a(p^e) = p^(10e)-p^(10(e-1)).
Dirichlet generating function: zeta(s-10)/zeta(s). - Ralf Stephan, Jul 04 2013
a(n) = n^10*Product_{distinct primes p dividing n} (1-1/p^10). - Tom Edgar, Jan 09 2015
Sum_{k=1..n} a(k) ~ n^11 / (11*zeta(11)). - Vaclav Kotesovec, Feb 07 2019
From Amiram Eldar, Oct 12 2020: (Start)
lim_{n->oo} (1/n) * Sum_{k=1..n} a(k)/k^10 = 1/zeta(11).
Sum_{n>=1} 1/a(n) = Product_{p prime} (1 + p^10/(p^10-1)^2) = 1.0009955309... (End)
MAPLE
f:= n -> n^10*mul(1-1/p^10, p=numtheory:-factorset(n)):
map(f, [$1..30]); # Robert Israel, Jan 09 2015
MATHEMATICA
JordanJ[n_, k_] := DivisorSum[n, #^k*MoebiusMu[n/#] &]; f[n_] := JordanJ[n, 10]; Array[f, 21]
f[p_, e_] := p^(10*e) - p^(10*(e-1)); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Oct 12 2020 *)
PROG
(PARI) for(n=1, 100, print1(sumdiv(n, d, d^10*moebius(n/d)), ", "))
CROSSREFS
KEYWORD
easy,nonn,mult
AUTHOR
Benoit Cloitre, Apr 05 2002
STATUS
approved